Chứng minh điều ngược lại đúng tức là. Cho a,b,c>0 thỏa \(b+c=2a\) thì \(\sqrt{b+1}+\sqrt{c+1}\le2\sqrt{a+1}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT=\left(\sqrt{b+1}+\sqrt{c+1}\right)^2\)

\(\le​\left(1+1\right)\left(b+1+c+1\right)\)

\(=2\left(b+c+2\right)\le4\left(a+1\right)=VP\)

\(\Rightarrow\left(\sqrt{b+1}+\sqrt{1+c}\right)^2\le4\left(a+1\right)\)

\(\Rightarrow\sqrt{b+1}+\sqrt{1+c}\le\sqrt{4\left(a+1\right)}=2\sqrt{a+1}\)

BĐT cuối đúng hay ta có ĐPCM

Chứng minh điều ngược lại đúng, tức là :Cho a,b,c>0 thỏa \(b+c=2a\) thì \(\sqrt{b+1}+\sqrt{c+1}\le2\sqrt{a+1}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{b+1}+\sqrt{c+1}\right)^2\)

\(\le\left(1+1\right)\left(b+1+c+1\right)\)

\(=2\left(b+c+2\right)=2\left(2a+2\right)\)

\(=4\left(a+1\right)=2^2\sqrt{\left(a+1\right)^2}=VP^2\)

Vì \(VT^2\le VP^2\Rightarrow VT\le VP\)

BĐT kia đúng nên ta có ĐPCM

sr bn mk tưởng chưa gửi dc nên gửi lại, Sorry

a) \(BĐT\Leftrightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)

\(\Leftrightarrow\sqrt{\frac{c\left(a-c\right)}{ab}}+\sqrt{\frac{c\left(b-c\right)}{ab}}\le1\)

\(\Leftrightarrow\sqrt{\frac{c}{b}\left(1-\frac{c}{a}\right)}+\sqrt{\frac{c}{a}\left(1-\frac{c}{b}\right)}\le1\)

Áp dụng AM-GM:\(VT\le\frac{1}{2}\left(\frac{c}{b}+1-\frac{c}{a}+\frac{c}{a}+1-\frac{c}{b}\right)=1\left(đpcm\right)\)

Dấu = xảy ra khi (a+b).c=ab

b) \(2+b+c+2+b+c\ge2\sqrt{\left(b+1\right)\left(c+1\right)}+2+b+c=\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge4\left(1+a\right)\)

\(\Leftrightarrow b+c\ge2a\)

cau a) dung cosi

\(\sqrt{c\left(a-c\right)}\le\frac{a-c+c}{2}\) ap dung cosi cho hai so c va a-c

tuong tu voi cac so khac

\(BT\le\frac{a-c+c}{2}+\frac{b-c+c}{2}-\frac{a+b}{2}\)(bt la VT cua de)

=> DPCM

b)

dung cosi nhu cau a

lam nhanh luon

\(\sqrt{1+b}\ge\frac{b+1+1}{2}\)

tuong tu

\(BT\ge\frac{b+2}{2}+\frac{c+2}{2}\ge a+2\)

<=> b+c>=2a

đánh dấu *

\(\sqrt{1+b}+\sqrt{1+c}=2\sqrt{1+a}\) (1)

⇒ \(b+c+2+2\sqrt{\left(1+b\right)\left(1+c\right)}=4a+4\)

⇒ \(b+c=4a+2-2\sqrt{\left(1+b\right)\left(1+c\right)}\)

Từ (1) ta lại có: \(2\sqrt{1+a}=\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt[4]{\left(1+b\right)\left(1+c\right)}\)

⇒ \(1+a\ge\sqrt{\left(1+b\right)\left(1+c\right)}\) ⇒

\(b+c=4a+2-2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4a+2-2\left(1+a\right)=2a\)

Vậy \(b+c\ge2a\), "=" xảy ra khi \(b=c=a\)

Ta có 

\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{b}+\frac{1}{a}\right)\)

                                                                \(\ge3+2a.\frac{1}{\sqrt{bc}}+2b.\frac{1}{\sqrt{ac}}+2c.\frac{1}{\sqrt{ab}}\)

Mà \(abc\le1\)

=> \(VT\ge3+2a\sqrt{a}+2b\sqrt{b}+2c\sqrt{c}=VP\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1